CN103401456A - Dual-modulating wave dual-carrier modulation method for voltage type three-level neutral point clamped converter - Google Patents

Abstract

The invention discloses a dual-modulating wave dual-carrier modulation method for a voltage type three-level neutral point clamped converter. The method comprises the following steps of (1) obtaining three phases of modulating waves according to a carrier PWM (pulse width modulation) method; (2) decomposing the modulating wave Ux of each phase into an upper modulating wave Uxp and a lower modulating wave Uxn; and (3) comparing the upper and lower modulating waves Uxp and Uxn with carriers, and outputting a final PWM pulse to the voltage type three-level neutral point clamped converter. According to the method, the upper and lower modulating waves Uxp and Uxn are solved according to the principle that average neutral point current in each carrier cycle is 0, and influence on the balance of direct current capacitance and voltage is avoided.

Description

The two carrier modulating methods of voltage-type three level neutral-point-clamped current transformer double modulation ripples

[technical field]

The present invention relates to electric and electronic technical field, particularly a kind of voltage-type three level neutral-point-clamped current transformer modulator approaches.

[background technology]

In recent years, three level neutral point clamped inverters are more and more extensive in the application in mesohigh field, and technology is also more and more ripe, still, still is being left to be desired aspect its intrinsic midpoint potential equilibrium problem.

The modulator approach of three-level inverter has Carrier-based PWM method and space vector PWM (space vector PWM, SVPWM) method.

SVPWM, because voltage utilization is high, vector is selected flexibly, be subject to welcoming widely, but this algorithm is complicated, amount of calculation is large.On the contrary, basic Carrier-based PWM has advantages of that but algorithm is simple, is easy to realize, yet, because three level NPC inverters need to be controlled the midpoint potential balance, and the method that generally adopts residual voltage to inject, this has brought following deficiency to Carrier-based PWM again: 1) calculating of the exact value of required injection residual voltage is very complicated, and this has offset the simple advantage of main carrier PWM algorithm; 2) under part modulation degree and power factor, the midpoint potential low-frequency fluctuation still can't be eliminated fully, and this not only needs to increase the direct current node capacitor, also can increase the low-frequency harmonics of inversion output.

In fact, by inject suitable residual voltage in modulating wave, Carrier-based PWM also can obtain the same high voltage utilization with SVPWM., if can further overcome the deficiency of Carrier-based PWM aspect the control of above-mentioned midpoint potential balance, will promote its application in three-level inverter, and further improve the performance of inverter.

[summary of the invention]

The two carrier modulating methods of voltage-type three level neutral-point-clamped current transformer double modulation ripples that the object of the present invention is to provide a kind of complete modulation degree, total digitalization, DC voltage balance do not impacted.

To achieve these goals, the technical solution used in the present invention is:

The two carrier modulating methods of voltage-type three level neutral-point-clamped current transformer double modulation ripples comprise the following steps:

(1), according to the Carrier-based PWM modulator approach, obtain the three-phase modulations ripple:

$\left\{\begin{array}{c}{U}_a=M\sin \left(\mathrm{\ωt}\right)\\ U_b=M\sin (\mathrm{\ωt}-2\mathrm{\π}/3)\\ U_c=M\sin (\mathrm{\ωt}+2\mathrm{\π}/3)\end{array}\right.---\left(1\right)$

Wherein M is modulation degree, and ω is the first-harmonic angular frequency;

(2) with each phase modulating wave U _xBe decomposed into modulating wave U _xpWith modulated ripple U _xnUpper modulating wave U _xpWith modulated ripple U _xnFormula as follows:

$\left\{\begin{array}{c}{U}_{\mathrm{xp}}=\frac{U_x+U_z+1-k}{2}(0\≤U_{\mathrm{xp}}\≤1)\\ U_{\mathrm{xn}}=\frac{U_x+U_z-1+k}{2}(-1\≤U_{\mathrm{xn}}\≤0)\end{array}\right.(x=a,b,c)---\left(2\right)$

K=k in formula (2) ₁Or k ₂Work as k=k ₁The time, U _Z=U _Z1; Work as k=k ₂The time, U _Z=U _Z2;

(3) k is brought in formula (2) into the upper modulating wave U that obtains every phase _xpWith modulated ripple U _xn, with upper modulating wave U _xp, modulated ripple U _xnCompare the final pwm pulse of rear output to voltage-type three level neutral-point-clamped current transformers with carrier wave.

, with respect to prior art, the invention has the beneficial effects as follows: the voltage-type three level carrier modulation strategies that the invention provides a kind of complete modulation degree, total digitalization, DC voltage balance do not impacted, the method realizes simple, effect is good.

[description of drawings]

Fig. 1 is three level diode clamp type current transformer main circuit topology figure;

Fig. 2 is the three-phase modulations ripple (M=1) of traditional three level diode clamp type current transformer carrier modulation strategies;

Fig. 3 is k=k ₁, U _z=U _z1The time the present invention three level diode clamp type current transformer double modulation ripples two carrier modulating methods three-phase modulations ripple (M=1);

Fig. 4 is k=k ₂, U _z=U _z2The time the present invention three level diode clamp type current transformer double modulation ripples two carrier modulating methods three-phase modulations ripple (M=1).

[embodiment]

The invention will be further described below in conjunction with accompanying drawing.

Provided as shown in Figure 1 the schematic diagram of diode clamp type three-level current transformer in the present invention, comprised that (if the three-level inverter structure, three-phase alternating current partly is load to the three-phase alternating current part; If the devices such as three level rectifiers, static reacance generator, three-phase alternating current partly adds three-phase AC flat wave reactor for alternating current source), the external part of three level DC side (if three-level inverter structure, the external part of DC side is direct voltage source, this DC source can be practical power, also can be the DC source that obtains by the AC power rectification; If three level rectifier structures, the external part of DC side is load; If three level static reacance generators, DC side is without external part), three level NPC current transformer main circuit parts, voltage sensor, current sensor, AD conversion chip and digital processing unit, wherein, voltage sensor senses three-phase alternating current part voltage and each capacitance voltage of DC side, current sensor detect each phase current of AC, voltage sensor is connected the AD conversion chip and is connected with digital processing unit with current sensor, digital processing unit and five-way are crossed corresponding drive circuit and controlled the switch of each power device in three-level current transformer.

The concrete steps of the two carrier modulating methods of voltage-type three level neutral-point-clamped current transformer double modulation ripples of the present invention are as follows:

(1) according to the Carrier-based PWM modulator approach, obtain corresponding three-phase modulations ripple, wherein M is modulation degree, ω is the first-harmonic angular frequency;

$\left\{\begin{array}{c}{U}_a=M\sin \left(\mathrm{\ωt}\right)\\ U_b=M\sin (\mathrm{\ωt}-2\mathrm{\π}/3)\\ U_c=M\sin (\mathrm{\ωt}+2\mathrm{\π}/3)\end{array}\right.---\left(1\right)$

(2) with each phase modulating wave U _x(x=a, b, c) is decomposed into modulating wave U _xpWith modulated ripple U _xn, be 0 principle according to the median average electric current in each carrier cycle, solve modulating wave U _xpWith modulated ripple U _xn(wherein k is the function of M and ω t, U _zFor the zero-sequence component that superposes in the three-phase modulations ripple);

$\left\{\begin{array}{c}{U}_{\mathrm{xp}}=\frac{U_x+U_z+1-k}{2}(0\≤U_{\mathrm{xp}}\≤1)\\ U_{\mathrm{xn}}=\frac{U_x+U_z-1+k}{2}(-1\≤U_{\mathrm{xn}}\≤0)\end{array}\right.(x=a,b,c)---\left(2\right)$

Provide its concrete solution procedure in the present invention:

Stack zero-sequence component U in traditional three-phase modulations ripple _zCan improve the voltage utilization of whole three level system, new modulating wave is:

U _x'＝U _x+U _z(x＝a,b,c) （3）

Definition according to upper modulating wave and modulated ripple:

U _xp+U _xn＝U _x+U _z(x＝a,b,c) （4）

Definition d _xoBe the duty ratio that in a carrier cycle, x is connected to dc bus capacitor mid point O, have:

d _xo＝1+U _xn-U _xp （5）

Want to make the carrier modulation strategy in each switch periods on dc capacitor voltage all without the impact, should guarantee:

Σ(1+U _xn-U _xp)·i _x≡0 （6）

Due to i _xWith U _xBetween phase difference fixing, a practical solution is:

1+U _an-U _ap＝1+U _bn-U _bp＝1+U _cn-U _cp （7）

And need to meet:

$\left\{\begin{array}{c}{U}_{\mathrm{xp}\max }={-U}_{\mathrm{xn}\min }\\ U_{\mathrm{xp}\min }=U_{\mathrm{xn}\max }=0\end{array}\right.(x=a,b,c)---\left(8\right)$

Order:

1+U _xn-U _xp＝k （9）

Have:

$\left\{\begin{array}{c}{U}_{\mathrm{xp}}=\frac{U_x+U_z+1-k}{2}\\ U_{\mathrm{xn}}=\frac{U_x+U_z-1+k}{2}\end{array}\right.---\left(10\right)$

Calculate k value and U according to the principle that reduces the modulation algorithm switching loss _z

Formula has k and U in (10) _zTwo variablees,, in order to reduce the switching loss of system, can make U in a period of time _xpFor being worth most (0 or 1), U _xn, for being worth most (0 or-1), solve:

$\left\{\begin{array}{c}{k}_{\mathrm{xp}\left(1\right)}=U_x+U_z-1\\ k_{\mathrm{xp}\left(0\right)}=U_x+U_z+1\\ k_{\mathrm{xn}\left(0\right)}={-U}_x-U_z+1\\ k_{\mathrm{xn}(-1)}={-U}_x-U_z-1\end{array}\right.(x=a,b,c)---\left(11\right)$

K in following formula _{Xy (j)}, x=a, b, c, y=p, n, j=1,0 ,-1, expression makes x(a, b, c) and y (p, the n) modulating wave of phase is j(1,0 ,-1) corresponding k separates during value.

There is following constraints in following formula:

$\left\{\begin{array}{c}{k}_{\mathrm{xp}\left(1\right)}={-k}_{\mathrm{xn}\left(0\right)},k_{\mathrm{xp}\left(0\right)}={-k}_{\mathrm{xn}(-1)},(x=a,b,c)\\ k_{\mathrm{xz}\left(j\right)}\≠k_{\mathrm{yz}\left(j\right)},(x,y=a,b,c,x\≠y,z=p,n,j=\mathrm{1,0},-1)\\ k_{\mathrm{xp}\left(1\right)}\≠k_{\mathrm{xp}\left(0\right)},k_{\mathrm{xn}\left(0\right)}\≠k_{\mathrm{xn}(-1)},(x=a,b,c)\end{array}\right.---\left(12\right)$

Owing to having U in above formula _zThis unknown number, can two k of simultaneous _{Xy (j)}Solve U _z:

$U_{\mathrm{zl}}=\left\{\begin{array}{c}1-U_x,k_{\mathrm{xp}\left(1\right)}=k_{\mathrm{xn}\left(0\right)}\\ {-U}_x,k_{\mathrm{xp}\left(1\right)}=k_{\mathrm{xn}(-1)}{\mathrm{ork}}_{\mathrm{xp}\left(0\right)}=k_{\mathrm{xn}\left(0\right)}\\ -1-U_x,k_{\mathrm{xp}\left(0\right)}=k_{\mathrm{xn}(-1)}\end{array}\right.(x=a,b,c)---\left(13\right)$

${\mathrm{<figure-callout\; id="2"\; label="U\; z"\; filenames="HDA00003486536600021.png,HDA00003486536600022.png"\; state="\{\{state\}\}">U</figure-callout>}}_z=\left\{\begin{array}{c}1-\frac{U_x+{\mathrm{<figure-callout\; id="2"\; label="U\; y"\; filenames="HDA00003486536600021.png,HDA00003486536600022.png"\; state="\{\{state\}\}">U</figure-callout>}}_y}{2},(k_{\mathrm{xp}}=k_{\mathrm{yn}\left(0\right)}{\mathrm{ork}}_{\mathrm{xn}\left(0\right)}=k_{\mathrm{yp}\left(1\right)})\\ -\frac{U_x+{\mathrm{<figure-callout\; id="2"\; label="U\; y"\; filenames="HDA00003486536600021.png,HDA00003486536600022.png"\; state="\{\{state\}\}">U</figure-callout>}}_y}{2},\left(\begin{array}{c}{k}_{\mathrm{xp}\left(1\right)}=k_{\mathrm{yn}(-1)}{\mathrm{ork}}_{\mathrm{xp}\left(0\right)}=k_{\mathrm{yn}\left(0\right)}\\ {\mathrm{ork}}_{\mathrm{xn}\left(0\right)}=k_{\mathrm{yp}\left(0\right)}{\mathrm{ork}}_{\mathrm{xn}(-1)}=k_{\mathrm{yp}\left(1\right)}\end{array}\right)\\ -1-\frac{U_x+{\mathrm{<figure-callout\; id="2"\; label="U\; y"\; filenames="HDA00003486536600021.png,HDA00003486536600022.png"\; state="\{\{state\}\}">U</figure-callout>}}_y}{2},(k_{\mathrm{xp}\left(0\right)}=k_{\mathrm{yn}(-1)}{\mathrm{ork}}_{\mathrm{xn}(-1)}=k_{\mathrm{yp}\left(0\right)})\end{array}\right.(x,y=a,b,c,x\&NotEqual;y)---\left(14\right)$

U _zNeed to meet with the k value:

$\left\{\begin{array}{c}0\&le;U_{\mathrm{xp}}=\frac{U_x+U_z+1-\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HDA00003486536600021.png,HDA00003486536600022.png"\; state="\{\{state\}\}">k</figure-callout>}}{2}\&le;1\\ -1\&le;U_{\mathrm{xn}}=\frac{U_x+U_z-1+\mathrm{<figure-callout\; id="2"\; label="k"\; filenames="HDA00003486536600021.png,HDA00003486536600022.png"\; state="\{\{state\}\}">k</figure-callout>}}{2}\&le;0\end{array}\right.---\left(15\right)$

Meet the U of following formula _zWith the scope of the solution of k and corresponding ω t thereof be:

K in following formula ₁/ U _z1And k ₂/ U _z2These two groups of solutions can make d _xoPerseverance is 0, and this result finally can make three level modulation deteriorate to two level modulation, need give up, and only has following solution in the historical facts or anecdotes border:

$\left\{\begin{array}{c}k=\&PlusMinus;[1+(U_b-U_c)/2],U_z=-(U_b+U_c)/2,\mathrm{\&omega;t}\&Element;[0,\mathrm{\&pi;}/6]\&cup;[11\mathrm{\&pi;}/\mathrm{6,2}\mathrm{\&pi;}]\\ k=\&PlusMinus;[1+(U_b-U_a)/2],U_z=-(U_a+U_b)/2,\mathrm{\&omega;t}\&Element;[\mathrm{\&pi;}/6,\mathrm{\&pi;}/2]\\ k=\&PlusMinus;[1+(U_c-U_a)/2],U_z=-(U_c+U_a)/2,\mathrm{\&omega;t}\&Element;[\mathrm{\&pi;}/\mathrm{2,5}\mathrm{\&pi;}/6]\\ k=\&PlusMinus;[1+(U_c-U_b)/2],U_z=-(U_b+U_c)/2,\mathrm{\&omega;t}\&Element;[5\mathrm{\&pi;}/6,7\mathrm{\&pi;}/6,]\\ k=\&PlusMinus;[1+(U_a-U_b)/2],U_z=-(U_a+U_b)/2,\mathrm{\&omega;t}\&Element;[7\mathrm{\&pi;}/\mathrm{6,3}\mathrm{\&pi;}/2]\\ k=\&PlusMinus;[1+(U_a-U_c)/2],U_z=-(U_c+U_a)/2,\mathrm{\&omega;t}\&Element;[3\mathrm{\&pi;}/\mathrm{2,11}\mathrm{\&pi;}/6]\end{array}\right.---\left(17\right)$

K in following formula and U _zFor piecewise function, k has two solutions, U in each π/3 zones _zA solution is arranged in each π/3 zones, have 2 ⁶=64 solutions, most solutions wherein all make U _xpAnd U _xnDiscontinuous, work as U _xpAnd U _xnConsecutive hours has:

$\left\{\begin{array}{c}{U}_{\mathrm{xy}(\frac{\mathrm{\&pi;}}{6}-)}=U_{\mathrm{xy}(\frac{\mathrm{\&pi;}}{6}+)}\\ U_{\mathrm{xy}(\frac{\mathrm{\&pi;}}{2}-)}=U_{\mathrm{xy}(\frac{\mathrm{\&pi;}}{2}+)}\\ U_{\mathrm{xy}(\frac{5\mathrm{\&pi;}}{6}-)}=U_{\mathrm{xy}(\frac{5\mathrm{\&pi;}}{6}+)}\\ U_{\mathrm{xy}(\frac{7\mathrm{\&pi;}}{6}-)}=U_{\mathrm{xy}(\frac{7\mathrm{\&pi;}}{6}+)}\\ U_{\mathrm{xy}(\frac{3\mathrm{\&pi;}}{2}-)}=U_{\mathrm{xy}(\frac{3\mathrm{\&pi;}}{2}+)}\\ U_{\mathrm{xy}(\frac{11\mathrm{\&pi;}}{6}-)}=U_{\mathrm{xy}(\frac{11\mathrm{\&pi;}}{6}+)}\end{array}\right.(x=a,b,c,y=p,n)---\left(18\right)$

, according to following formula, solve the upper modulating wave U that makes the x phase _xp(modulated ripple U _xn) continuous solution is:

(3) k is brought into the U of modulating wave up and down that obtains the every phase of new type of modulation strategy in formula (2) _xpAnd U _xn, will up and down modulating wave U _xpAnd U _xnCompare the final pwm pulse of rear output to voltage-type three level neutral-point-clamped current transformers with carrier wave.

Fig. 3 and Fig. 4 are respectively k and Uz is respectively k ₁, U _z1And k ₂, U _z2The time three-phase up and down modulating wave U _xpAnd U _xn, compared to traditional single-phase dull ripple strategy, the new type of modulation strategy that the present invention proposes increases degree of freedom k and U _z, by k and U _zReasonable value just can so that the balance of modulation strategy alignment voltage without impact, k and U _zDerivation comparatively complicated, but result is very simple, realizes simple and easy to doly, is not subjected to the impact of modulation degree and load simultaneously.But need to prove, the accent strategy that the present invention proposes can increase the switching loss of power device, with traditional carrier wave SPWM strategy, compares, and has increased the devices switch loss of 1/3 times.

Claims (1)

1. the two carrier modulating methods of voltage-type three level neutral-point-clamped current transformer double modulation ripples, is characterized in that, comprises the following steps:

(1), according to the Carrier-based PWM modulator approach, obtain the three-phase modulations ripple:

$\left\{\begin{array}{c}{U}_a=M\sin \left(\mathrm{\&omega;t}\right)\\ U_b=M\sin (\mathrm{\&omega;t}-2\mathrm{\&pi;}/3)\\ U_c=M\sin (\mathrm{\&omega;t}+2\mathrm{\&pi;}/3)\end{array}\right.---\left(1\right)$

Wherein M is modulation degree, and ω is the first-harmonic angular frequency;

(2) with each phase modulating wave U _xBe decomposed into modulating wave U _xpWith modulated ripple U _xnUpper modulating wave U _xpWith modulated ripple U _xnFormula as follows:

$\left\{\begin{array}{c}{U}_{\mathrm{xp}}=\frac{U_x+U_z+1-k}{2}(0\&le;U_{\mathrm{xp}}\&le;1)\\ U_{\mathrm{xn}}=\frac{U_x+U_z-1+k}{2}(-1\&le;U_{\mathrm{xn}}\&le;0)\end{array}\right.(x=a,b,c)---\left(2\right)$

K=k in formula (2) ₁Or k ₂Work as k=k ₁The time, U _Z=U _Z1; Work as k=k ₂The time, U _Z=U _Z2;

(3) k is brought in formula (2) into the upper modulating wave U that obtains every phase _xpWith modulated ripple U _xn, with upper modulating wave U _xp, modulated ripple U _xnCompare the final pwm pulse of rear output to voltage-type three level neutral-point-clamped current transformers with carrier wave.

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